Subcontests
(5)2000 BAMO p3 product (x_i+ 1/x_i) >= product (x_i+ 1/x_{i+1})
Let x1,x2,...,xn be positive numbers, with n≥2. Prove that
(x1+x11)(x2+x21)...(xn+xn1)≥(x1+x21)(x2+x31)...(xn−1+xn1)(xn+x11) 2000 BAMO p2 concurrent circles wanted
Let ABC be a triangle with D the midpoint of side AB,E the midpoint of side BC, and F the midpoint of side AC. Let k1 be the circle passing through points A,D, and F, let k2 be the circle passing through points B,E, and D, and let k3 be the circle passing through C,F, and E. Prove that circles k1,k2, and k3 intersect in a point. 2000 BAMO p5 Alice plays solitaire on a 20x20 chessboard, 59 of 400 coins
Alice plays the following game of solitaire on a 20×20 chessboard.
She begins by placing 100 pennies, 100 nickels, 100 dimes, and 100 quarters on the board so that each of the 400 squares contains exactly one coin. She then chooses 59 of these coins and removes them from the board.
After that, she removes coins, one at a time, subject to the following rules:
- A penny may be removed only if there are four squares of the board adjacent to its square (up, down, left, and right) that are vacant (do not contain coins). Squares “off the board” do not count towards this four: for example, a non-corner square bordering the edge of the board has three adjacent squares, so a penny in such a square cannot be removed under this rule, even if all three adjacent squares are vacant.
- A nickel may be removed only if there are at least three vacant squares adjacent to its square. (And again, “off the board” squares do not count.)
- A dime may be removed only if there are at least two vacant squares adjacent to its square (“off the board” squares do not count).
- A quarter may be removed only if there is at least one vacant square adjacent to its square (“off the board” squares do not count).
Alice wins if she eventually succeeds in removing all the coins. Prove that it is impossiblefor her to win.